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Black_prince [1.1K]
3 years ago
8

What the answer to ⅓( 6x-9)

Mathematics
1 answer:
Vikki [24]3 years ago
5 0
The answer is 2x-3 hope this helpsss
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Which exponential functions have been simplified correctly check all that apply
Fittoniya [83]
Principle: Law of Exponents - Combination of product to a power & power to a power. The first is when raising a product of two integers to a power, the power is distributed to each factor. In equation it is,

(xy)^a = (x^a)(y^a)

The latter is when raising the base with a power to a power, the base will remain the same and the powers will be multiplied. In equation it is, 
(x^a)(x^b) = x^ab

Check:


f(x) = 5*(16)^.33x = 5*(8*2)^0.33x = 5*(8^0.33x)(2^0.33x) = 5*(2^x)*(2^0.33x) = 5*(2^1.33x)

f(x) = 2.3*(8^0.5x) = 2.3*(4*2)^0.5x = 2.3*(2^x)(2^0.5x) = 2.3*(2^1.5x)

f(x) = 81^0.25x = 3^x

f(x) = 0.75*(9*3)^0.5x = 0.75*(3^x)*(3^0.5x) = 0.75*3^1.5x

f(x) = 24^0.33x = (8*3)^0.33x = (2^x)*(3^0.33x)


Therefore, the answer is third equation.

<em>ANSWER: f(x) = 81^0.25x = 3^x</em>
5 0
3 years ago
Read 2 more answers
A baker buys 19 apples of two different varieties to make pies. The total cost of the apples is $5.10. Granny Smith apples cost
Natasha2012 [34]

The baker bought 7 gala apples, and 12 granny smiths apples

<em><u>Solution:</u></em>

Let "x" be the number of gala apples bought

Let "y" be the number of granny smith apples bought

Cost of 1 gala apple = $ 0.30

Cost of 1 granny smith apple = $ 0.25

<em><u>A baker buys 19 apples of two different varieties to make pies</u></em>

Therefore,

number of gala apples bought + number of granny smith apples bought = 19

x + y = 19 --------- eqn 1

<em><u>The total cost of the apples is $5.10</u></em>

Therefore, we can frame a equation as:

number of gala apples bought x Cost of 1 gala apple + number of granny smith apples bought x Cost of 1 granny smith apple = 5.10

x \times 0.30 + y \times 0.25 = 5.10

0.3x + 0.25y = 5.1 -------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

x = 19 - y ---------- eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

0.3(19 - y) + 0.25y = 5.1

5.7 - 0.3y + 0.25y = 5.1

5.7 - 0.05y = 5.1

0.05y = 5.7 - 5.1

0.05y = 0.6

y = 12

<em><u>Substitute y = 12 in eqn 3</u></em>

x = 19 - 12

x = 7

Thus the baker bought 7 gala apples, and 12 granny smiths apples

6 0
3 years ago
Find the 13th term of the sequence 4, 8, 16, 32, .
topjm [15]

Answer:

The thirteen number will be 16384.

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
9. Write an equation for the line that is parallel to the given line and that passes through the given point.
timurjin [86]

Answer:

  y -11 = 2(x -3)

Step-by-step explanation:

The slope of the given line is the x-coefficient: 2. The parallel line will have the same slope.

When you know the slope and a point on the line, it is convenient to use the point-slope form of the equation of a line:

  y -k = m(x -h) . . . . . . line with slope m through point (h, k)

Your line's equation is ...

  y -11 = 2(x -3)

7 0
3 years ago
Here are the endpoints of the segments BC, FG, and JK.<br> B, −67
yulyashka [42]

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{[-4 - (-6)]^2 + [4 - 7]^2}\implies BC=\sqrt{(-4+6)^2+(-3)^2} \\\\\\ BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}

F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}

JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill

4 0
2 years ago
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